• 【紫书】Urban Elevations UVA


    题意:给你俯视图,要求依次输出正视图中可以看到的建筑物

    题解:任意相邻的x间属性相同,所以离散化。

    坑:unique只能对数组用。下标易错

      list不能找某元素的next。用了个很麻烦的处理

    数组:

    #define _CRT_SECURE_NO_WARNINGS
    #include<cstring>
    #include<cctype>
    #include<cstdlib>
    #include<cmath>
    #include<cstdio>
    #include<string>
    #include<stack>
    #include<ctime>
    #include<list>
    #include<set>
    #include<map>
    #include<queue>
    #include<vector>
    #include<sstream>
    #include<iostream>
    #include<functional>
    #include<algorithm>
    #include<memory.h>
    //#define INF 0x3f3f3f3f
    #define eps 1e-6
    #define pi acos(-1.0)
    #define e exp(1.0)
    #define rep(i,t,n)  for(int i =(t);i<=(n);++i)
    #define per(i,n,t)  for(int i =(n);i>=(t);--i)
    #define mp make_pair
    #define pb push_back
    #define mmm(a,b) memset(a,b,sizeof(a))
    //std::ios::sync_with_stdio(false);
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    void smain();
    #define ONLINE_JUDGE
    int main() {
        //ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        long _begin_time = clock();
    #endif
        smain();
    #ifndef ONLINE_JUDGE
        long _end_time = clock();
        printf("time = %ld ms.", _end_time - _begin_time);
    #endif
        return 0;
    }
    const int maxn = 1e2 + 5;
    struct building {
        int id;
        double x, y, w, d, h;
        bool operator<(const building& rhs)const {
            return x < rhs.x || (x == rhs.x&&y < rhs.y);
        }
    }b[maxn];
    int n;
    double x[maxn * 2];
    bool cover(int i, double mx) {
        return b[i].x<mx&&b[i].x + b[i].w>mx;
    }
    bool visible(int i, double mx) {
        if (!cover(i, mx))return false;
        rep(k, 1, n)
            if (b[k].y < b[i].y&&b[k].h >= b[i].h&&cover(k, mx))return false;
        return true;
    }
    
    void Run() {
    
    }
    
    void smain() {
    
        int kase = 0;
        while (cin >> n) {
            if (n == 0)break;
            rep(i, 1, n)
            {
                cin >> b[i].x >> b[i].y >> b[i].w >> b[i].d >> b[i].h;
                x[i * 2] = b[i].x; x[i * 2 + 1] = b[i].x + b[i].w;
                b[i].id = i;
            }
            sort(b + 1, b + 1 + n);
            sort(x + 2, x + 2 + n * 2);
            int m = unique(x + 2, x + 2 + n * 2) - x-2;
            if (kase++)cout << endl;
            printf("For map #%d, the visible buildings are numbered as follows:
    %d", kase, b[1].id);
            rep(i, 2, n) {
                bool vis = false;
                rep(j, 2,m ) if (visible(i, (x[j] + x[j + 1]) / 2)) { vis = true; break; }
                if (vis)cout << ' ' << b[i].id;
            }
            cout << endl;
    
    
    
        }
    }

    用list

    #define _CRT_SECURE_NO_WARNINGS
    #include<cstring>
    #include<cctype>
    #include<cstdlib>
    #include<cmath>
    #include<cstdio>
    #include<string>
    #include<stack>
    #include<ctime>
    #include<list>
    #include<set>
    #include<map>
    #include<queue>
    #include<vector>
    #include<sstream>
    #include<iostream>
    #include<functional>
    #include<algorithm>
    #include<memory.h>
    //#define INF 0x3f3f3f3f
    #define eps 1e-6
    #define pi acos(-1.0)
    #define e exp(1.0)
    #define rep(i,t,n)  for(int i =(t);i<=(n);++i)
    #define per(i,n,t)  for(int i =(n);i>=(t);--i)
    #define mp make_pair
    #define pb push_back
    #define mmm(a,b) memset(a,b,sizeof(a))
    //std::ios::sync_with_stdio(false);
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    void smain();
    #define ONLINE_JUDGE
    int main() {
        //ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        long _begin_time = clock();
    #endif
        smain();
    #ifndef ONLINE_JUDGE
        long _end_time = clock();
        printf("time = %ld ms.", _end_time - _begin_time);
    #endif
        return 0;
    }
    const int maxn = 1e2 + 5;
    struct building {
        int id;
        double x, y, w, d, h;
        bool operator<(const building& rhs)const {
            return x < rhs.x || (x == rhs.x&&y < rhs.y);
        }
    }b[maxn];
    int n;
    double x[maxn * 2];
    list<double> l;
    bool cover(int i, double mx) {
        return b[i].x<mx&&b[i].x + b[i].w>mx;
    }
    bool visible(int i, double mx) {
        if (!cover(i, mx))return false;
        rep(k, 1, n)
            if (b[k].y < b[i].y&&b[k].h >= b[i].h&&cover(k, mx))return false;
        return true;
    }
    
    void Run() {
    
    }
    
    void smain() {
    
        int kase = 0;
        while (cin >> n) {
            if (n == 0)break;
            rep(i, 1, n)
            {
                cin >> b[i].x >> b[i].y >> b[i].w >> b[i].d >> b[i].h;
                
                l.push_back(b[i].x), l.push_back(b[i].x + b[i].w);
                b[i].id = i;
            }
            sort(b + 1, b + 1 + n);
            
            l.sort();
            
            l.unique();
            if (kase++)cout << endl;
            printf("For map #%d, the visible buildings are numbered as follows:
    %d", kase, b[1].id);
            rep(i, 2, n) {
                bool vis = false;
                int last = l.front();
                for (auto t : l) if (t != l.front() && visible(i, (t + last) / 2)) { vis = true; break; }
                else last = t;
                if (vis)cout << ' ' << b[i].id;
            }
            cout << endl;
    
    
    
        }
    }
    成功的路并不拥挤,因为大部分人都在颓(笑)
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  • 原文地址:https://www.cnblogs.com/SuuT/p/8988914.html
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